Okay so I've been reviewing previous tests for my midterm exam and I came across a question that I'm not too sure how to solve (I got this answer wrong and didn't bother to correct it so I don't know the solution) :((adsbygoogle = window.adsbygoogle || []).push({});

The four legged table as shown has mass 30 kg. The table top has mass 20 kg and each leg is 2.5 kg. Find the centre of mass of the table and the angle at which the table will overturn. What minimum force applied to the upper edge of the table would put the table in a state of equilibrium and thus would tip with any larger force?

*note the table top has a width of 10 cm (couldn't draw it in)

|-----1.4 m-----|

_______________ _

| |

| |

| | 1.0 m

| | _

<---1.0 m--->

Okay so I think I figured out the first part. the center of mass of the table can be found by doing a torque calculation with the table rotated 90 degrees:

ƩTorque = 0

-Fgtabletop(d) + Fglegs(50-d) = 0

-196d + 4900 - 98d = 0

d = 16.666666

Therefore, Centre of mass = 95 cm - d = 95 - 16.66666 = 78.3333 cm = 0.78333 m

Now, the critical tipping angle can be calculated:

tan θ = (1/2width)/(centre of mass) = (0.5 m)/(0.783m)

θ = 32.6°

If someone could please verify my answer and help me with the last part regarding the minimum force applied to the upper edge it would be greatly appreciated. Thanks.

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# Critical Tipping Angle Question

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